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    (Original post by SeanFM)
    P(Z<1.96) = 0.975.

    Remember that with an a value when we want to find P(X<a) we do (a-mean)/standard deviation to turn a into z, and find P(Z<z).

    If a was 18 for example and we wanted to find P(X<18) then we would do P(Z<(18-15)/4) = P(Z<3/4).

    What I was trying to hint at was that each value for a has a value for z that we calcuate, and in that example (a-15)/4 = 3/4, and in your question it's ..... = 1.96.
    so
    a-15 / 4 = 1.96
    multiply by 4
    a-15 = 7.84
    add 15
    a = 22.84 ??
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    (Original post by Aty100)
    so
    a-15 / 4 = 1.96
    multiply by 4
    a-15 = 7.84
    add 15
    a = 22.84 ??
    :party2:Yes, well done

    If you can now see what I meant by 'working backwards' then you have done very well.
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    (Original post by SeanFM)
    :party2:Yes, well done

    If you can now see what I meant by 'working backwards' then you have done very well.
    FINALLY !

    Yess thanks soo much for your time !
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    (Original post by SeanFM)
    :party2:Yes, well done

    If you can now see what I meant by 'working backwards' then you have done very well.
    so for
    X~N ( 85 , 25)
    P ( X > a ) = 0.67

    would you do 1 - 0.67 because it is greater than and that is 0.33

    a - 85 / 5 = 0.33 ?
    a - 85 = 1.65
    a = 85.65
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    (Original post by Aty100)
    so for
    X~N ( 85 , 25)
    P ( X > a ) = 0.67

    would you do 1 - 0.67 because it is greater than and that is 0.33

    a - 85 / 5 = 0.33 ?
    a - 85 = 1.65
    a = 85.65
    Remember that you need to equate the transformation (Which is (a-mean)/standard deviation) to a z value, not the actual probability.
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    (Original post by SeanFM)
    Remember that you need to equate the transformation (Which is (a-mean)/standard deviation) to a z value, not the actual probability.
    so would I equal to 0.6293 ?
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    (Original post by Aty100)
    so would I equal to 0.6293 ?
    If that's the correct value from the tables, then yes.
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    (Original post by SeanFM)
    If that's the correct value from the tables, then yes.
    You're soo awesome! Thanks once again for your time and help. I appreciate it !
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    (Original post by Aty100)
    You're soo awesome! Thanks once again for your time and help. I appreciate it !
    No problem
 
 
 
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